A-cubical-block-is-held-stationary-against-a-rough-wall-by-applying-force-F-then-incorrect-statement-among-the-following-is-1-frictional-force-f-Mg-2-f-N-N-is-normal-reaction-3-F-does-no

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Question Number 22437 by Tinkutara last updated on 18/Oct/17

$$\mathrm{A}\mathrm{cubical}\mathrm{block}\mathrm{is}\mathrm{held}\mathrm{stationary}$$$$\mathrm{against}\mathrm{a}\mathrm{rough}\mathrm{wall}\mathrm{by}\mathrm{applying}\mathrm{force}$$$${}^{\prime }{\mathrm{F}}^{\prime }\mathrm{then}\mathit{i}\mathit{n}\mathit{c}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathrm{statement}\mathrm{among}$$$$\mathrm{the}\mathrm{following}\mathrm{is}$$$$\left(1\right)\mathrm{frictional}\mathrm{force},\mathrm{f}=\mathrm{Mg}$$$$\left(2\right)\mathrm{f}=\mathrm{N},\mathrm{N}\mathrm{is}\mathrm{normal}\mathrm{reaction}$$$$\left(3\right)\mathrm{F}\mathrm{does}\mathrm{not}\mathrm{apply}\mathrm{any}\mathrm{torque}$$$$\left(4\right)\mathrm{N}\mathrm{does}\mathrm{not}\mathrm{apply}\mathrm{any}\mathrm{torque}$$

Commented by Tinkutara last updated on 18/Oct/17

Commented by math solver last updated on 18/Oct/17

$$ithinkstatement2iswrongi.ef=N.$$

Commented by Tinkutara last updated on 18/Oct/17

$$\mathrm{No}.\mathrm{Wrong}\mathrm{answer}.$$

Commented by ajfour last updated on 18/Oct/17

$$onlystatement\left(4\right)isincorrect.$$$$\left(2\right)mightbecorrectdepending$$$$onvalueof\mu .$$

Commented by Tinkutara last updated on 18/Oct/17

$$\left(3\right)\mathrm{should}\mathrm{be}\mathrm{wrong}\mathrm{since}\mathrm{line}\mathrm{of}\mathrm{action}$$$$\mathrm{of}\mathrm{F}\mathrm{passes}\mathrm{through}\mathrm{COM}\mathrm{of}\mathrm{block}.\mathrm{Why}$$$$\mathrm{not}?$$

Commented by mrW1 last updated on 18/Oct/17

Commented by mrW1 last updated on 18/Oct/17

$$\mathrm{f}=\mathrm{Mg}$$$$\mathrm{N}=\mathrm{F}$$$$\mathrm{f}\times \mathrm{a}=\mathrm{N}\times \mathrm{e}$$$$\Rightarrow \mathrm{e}=\frac{\mathrm{f}\times \mathrm{a}}{\mathrm{N}}=\frac{\mathrm{Mga}}{\mathrm{F}}\ne 0$$$$\mathrm{F},\mathrm{Mg}\mathrm{apply}\mathrm{no}\mathrm{torque}.$$$$\mathrm{f},\mathrm{N}\mathrm{apply}\mathrm{torque}.$$$$\Rightarrow \left(4\right)\mathrm{is}\mathrm{wrong}$$

Commented by Tinkutara last updated on 18/Oct/17

$$\mathrm{Why}\mathrm{N}\mathrm{is}\mathrm{not}\mathrm{along}\mathrm{COM}?$$

Commented by mrW1 last updated on 18/Oct/17

$$\mathrm{we}\mathrm{have}4\mathrm{forces}.\mathrm{F}\mathrm{and}\mathrm{Mg}\mathrm{go}\mathrm{through}$$$$\mathrm{COM},\mathrm{f}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{go}\mathrm{through}\mathrm{COM}.$$$$\mathrm{If}\mathrm{N}\mathrm{also}\mathrm{goes}\mathrm{through}\mathrm{COM},\mathrm{then}$$$$\mathrm{the}\mathrm{block}\mathrm{would}\mathrm{rotate}\mathrm{due}\mathrm{to}\mathrm{the}$$$$\mathrm{torque}\mathrm{created}\mathrm{by}\mathrm{f}.\mathrm{That}\mathrm{means}\mathrm{N}$$$$\mathrm{must}\mathrm{create}\mathrm{a}\mathrm{torque}\mathrm{opposite}\mathrm{to}\mathrm{that}$$$$\mathrm{created}\mathrm{by}\mathrm{f}.$$

Commented by ajfour last updated on 18/Oct/17

$$thatswhy\left(3\right)iscorrect$$$$andwehavetochooseincorrect.$$$$NotorqueofFaboutCOMas$$$$itslineofactionpassesthrough$$$$COMofblock.$$$$LineofactionofnetNormal$$$$forcedoesshiftansappliesa$$$$torquethatcounterbalances$$$$torqueoffriction.$$

Commented by Tinkutara last updated on 18/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!\mathrm{Now}\mathrm{I}\mathrm{got}$$$$\mathrm{understood}.$$

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